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Difference between revisions of "2015 AMC 8 Problems/Problem 2"

(Add the solution, and the content in the See Also header.)
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</asy>
 
</asy>
  
−
==Solution==
+
==Solution 1==
  
 
Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into 8 equal parts, such as triangles <math>\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO</math>, etc.  These parts, since they are all equal, are <math>\frac{1}{8}</math> of the octagon each.  The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is <math>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.</math>
 
Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into 8 equal parts, such as triangles <math>\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO</math>, etc.  These parts, since they are all equal, are <math>\frac{1}{8}</math> of the octagon each.  The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is <math>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.</math>
 +
 +
==Solution 2==
 +
[asy]
 +
pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g;
 +
A=dir(45);
 +
B=dir(90);
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C=dir(135);
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D=dir(180);
 +
E=dir(-135);
 +
F=dir(-90);
 +
G=dir(-45);
 +
H=dir(0);
 +
O=(0,0);
 +
X=midpoint(A--B);
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a=midpoint(B--C);
 +
b=midpoint(C--D);
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c=midpoint(D--E);
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d=midpoint(E--F);
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e=midpoint(F--G);
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f=midpoint(G--H);
 +
g=midpoint(H--A);
 +
 +
fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));
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draw(A--B--C--D--E--F--G--H--cycle);
 +
 +
dot("<math>A</math>",A,dir(45));
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dot("<math>B</math>",B,dir(90));
 +
dot("<math>C</math>",C,dir(135));
 +
dot("<math>D</math>",D,dir(180));
 +
dot("<math>E</math>",E,dir(-135));
 +
dot("<math>F</math>",F,dir(-90));
 +
dot("<math>G</math>",G,dir(-45));
 +
dot("<math>H</math>",H,dir(0));
 +
dot("<math>X</math>",X,dir(135/2));
 +
dot("<math>O</math>",O,dir(0));
 +
draw(E--O--X);
 +
draw(B--F);
 +
draw(A--O);
 +
draw(D--H);
 +
draw(C--G);
 +
draw(a--e);
 +
draw(b--f);
 +
draw(c--g);
 +
draw(d--O);
 +
[/asy]
 +
 +
The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:50, 25 November 2015

Point $O$ is the center of the regular octagon $ABCDEFGH$, and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded?

$\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}$

[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("$A$",A,dir(45)); dot("$B$",B,dir(90)); dot("$C$",C,dir(135)); dot("$D$",D,dir(180)); dot("$E$",E,dir(-135)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-45)); dot("$H$",H,dir(0)); dot("$X$",X,dir(135/2)); dot("$O$",O,dir(0)); draw(E--O--X); [/asy]

Solution 1

Since octagon $ABCDEFGH$ is a regular octagon, it is split into 8 equal parts, such as triangles $\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO$, etc. These parts, since they are all equal, are $\frac{1}{8}$ of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is $\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.$

Solution 2

[asy] pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); a=midpoint(B--C); b=midpoint(C--D); c=midpoint(D--E); d=midpoint(E--F); e=midpoint(F--G); f=midpoint(G--H); g=midpoint(H--A);

fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle);

dot("$A$",A,dir(45)); dot("$B$",B,dir(90)); dot("$C$",C,dir(135)); dot("$D$",D,dir(180)); dot("$E$",E,dir(-135)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-45)); dot("$H$",H,dir(0)); dot("$X$",X,dir(135/2)); dot("$O$",O,dir(0)); draw(E--O--X); draw(B--F); draw(A--O); draw(D--H); draw(C--G); draw(a--e); draw(b--f); draw(c--g); draw(d--O); [/asy]

The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is $\boxed{\textbf{(D)}~\dfrac{7}{16}}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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